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Yahoo!Best ResponseYou've already chosen the best response.1
@apoorvk @satellite73 @A.Avinash_Goutham @ajprincess @AndrewNJ
 one year ago

apoorvkBest ResponseYou've already chosen the best response.1
okay, expand the two wholesquares, it's simple enough...
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.1
the options are 4cos^2(ab/2) 4 sin a/2 cos b/2
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.1
@apoorvk after expanding i got this then next..............
 one year ago

apoorvkBest ResponseYou've already chosen the best response.1
great, now.. cosAcosB + sinAsinB = cos(AB) Use this^
 one year ago

patmyer1Best ResponseYou've already chosen the best response.0
ugh, trig identities are a drag. Even after deriving them myself I simply CANNOT seem to memorize them
 one year ago

AndrewNJBest ResponseYou've already chosen the best response.1
spread it so we have (cosa)^2+(cosb)^2+2(cosa)(cosb)+(sina)^2+(sinb)^2+2(sina)(sinb) =2+2cos(ab)
 one year ago

AndrewNJBest ResponseYou've already chosen the best response.1
(cosa)^2+(sina)^2=1 (cosb)^2+(sinb)^2=1 (cosa)(cosb)+(sina)(sinb)=cos(ab)
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.1
ok @apoorvk so it is not over yet!!@
 one year ago

shivam_bhallaBest ResponseYou've already chosen the best response.2
Continuing from where @AndrewNJ left, \[2(1+\cos(ab))\] now we know that \[1 + \cos 2a = 2 \cos^2 a\] Now try :D
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.1
@shivam_bhalla did nt understand!
 one year ago

shivam_bhallaBest ResponseYou've already chosen the best response.2
= \[2(1+\cos 2\frac{(ab)}{2})\] Now we have a identity, \[1 + \cos 2x = 2 \cos^2 x\] in your case, \[x = \frac{ab}{2}\] Now try
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.1
i got it thxx @shivam_bhalla
 one year ago

Yahoo!Best ResponseYou've already chosen the best response.1
@shivam_bhalla plzz help http://openstudy.com/study?signup#/updates/4fd60c0be4b04bec7f170b7d
 one year ago
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